Re: Prove by induction that for every positive integer n

Thanks HallsofIvy,
Induction means a mathematical technique which is used to prove a statement, a formula or a theorem is true for every natural number.
Do I simply just sub in 1 for k into 2^k -2 which you have written after the last summation?
making it equal to 2 -2 = 0? and compare it with the earlier summation
I'm kinda not sure how to do the second part with the k+1 power added and break it down from there ....

Re: Prove by induction that for every positive integer n

Thanks HallsofIvy,
Induction means a mathematical technique which is used to prove a statement, a formula or a theorem is true for every natural number.
Do I simply just sub in 1 for k into 2^k -2 which you have written after the last summation?
making it equal to 2 -2 = 0? and compare it with the earlier summation
I'm kinda not sure how to do the second part with the k+1 power added and break it down from there ....

Re: Prove by induction that for every positive integer n

Could Anyone please give me an Idea of how I would go about solving this problem .?

Any advice would be appreciated cheers.

Let's approach this three ways. We want to prove that P(n) is true for every natural number.

The formal approach. Let S be the set for which every element s is a natural number such that P(s) is true. But S may be empty. So prove P(1) is true. Thus S is not empty. Let k be ANY element of S (there is at least one element in S). Because k in S, P(k) is true. So we we know that 1, k, and k + 1 are all natural numbers and that P(1) and P(k) are true. Using those five facts prove P(k+1) is true. Thus k + 1 in S. Therefore S = N.

That is the formal structure of a proof by weak mathematical induction.

The intuitive approach. If something is true for 1 and for the next number past a number for which that something is true, then it is true for 1, but it is also true for 1 + 1, the number next after 1 or 2, which makes it true for the number next after 2, or 3, which makes it true for 4 and then 5 and so on forever. The intuition is of an infinite number of dominoes falling.

The practical approach. The n = 1 step is usually easy. The challenge arises on the k to k + 1 step. There probably is no general way to attack all proofs by induction, but a very common practical approach when you are asked to equate two expressions is to form your expression in k + 1 and expand it into an expression involving k and known numbers without any mention of (k + 1).